First, let's calculate the horizontal component of the force applied by the boy:
[tex]\begin{gathered} F_x=F\cdot\cos30°\\ \\ F_x=80\cdot0.866\\ \\ F_x=69.28\text{ N} \end{gathered}[/tex]
Now, let's calculate the works:
a.
The work done by the boy will use only the horizontal component of the force applied by the boy:
[tex]\begin{gathered} W=F_x\cdot d\\ \\ W=69.28\cdot8\\ \\ W=554.24\text{ J} \end{gathered}[/tex]
b.
The work done by the friction force uses only the friction force:
[tex]\begin{gathered} W_f=F_f\cdot d\\ \\ W_f=-30\cdot8\\ \\ W_f=-240\text{ J} \end{gathered}[/tex]
(We use a negative force because it is against the movement direction).
c.
The net work done on the mower is:
[tex]\begin{gathered} W_{net}=W+W_f\\ \\ W_{net}=554.24-240\\ \\ W_{net}=314.24\text{ J} \end{gathered}[/tex]
d.
The change in kinetic energy is equal to the net work, so it is equal to 314.24 J.
e.
Using the kinetic energy formula, we have:
[tex]\begin{gathered} KE=\frac{mv^2}{2}\\ \\ 314.24=10\cdot\frac{v^2}{2}\\ \\ v^2=\frac{314.24}{5}\\ \\ v^2=62.85\\ \\ v=7.93\text{ m/s} \end{gathered}[/tex]
